the answer is: -1/2 but i keep on getting the wrong answer please help. Please include calculations and explaination of how u wrked it out thank you. :)
2007-01-03 06:48:54 · 5 answers · asked by izzy 1 in Science & Mathematics ➔ Mathematics
The geometric series
a + a*r + a*r^2 + a*r^3 + ...
has sum equal to a/(1-r)
In your case, a = 6 and the sum is 4.
4 = 6/(1-r)
1-r = 6/4 = 3/2
r = -1/2
2007-01-03 06:52:45 · answer #1 · answered by AnyMouse 3 · 1⤊ 0⤋
S=a/(1-r)
4=6/(1-r)
4(1-r)=6
4-4r=6
4-6=4r
4r=-2
r=-1/2
2007-01-03 15:23:00 · answer #2 · answered by Como 7 · 0⤊ 0⤋
The sum of an infinite geometric progression is:
S = a/(1-r), where a is the first term and r is the common ratio.
So 4= 6/(1-r)
4-4r=6
-4r = 6-4 = 2
r = -2/4 = -1/2
2007-01-03 15:39:48 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
Remember that the sum of an infinite series is determined by the formula
S = a1/(1 - r)
Where S is the sum and a1 is the first term.
Since S = 4, a1 = 6, it's just algebra.
4 = 6/(1 - r)
4(1 - r) = 6
4 - 4r = 6
-4r = 2
r = -1/2
Therefore the common ratio is -1/2
2007-01-03 14:53:37 · answer #4 · answered by Puggy 7 · 0⤊ 0⤋
4 = 6/(1-r)
Solve for r, r = -1/2
2007-01-03 14:57:57 · answer #5 · answered by sahsjing 7 · 0⤊ 0⤋